Prove $\lim_{x\to 1} \frac{x+2}{x^2+1}=\frac{3}{2}$ The title is quite clear. I am required to prove $\lim_{x\to 1} \frac{x+2}{x^2+1}=\frac{3}{2}$ using the epsilon-delta definition of the limit.

Given any $\varepsilon \gt 0$, there exists a $\delta =$
Such that $0 \lt \lvert x-1 \rvert \lt \delta \Rightarrow \lvert\frac{x+2}{x^2+1} -\frac{3}{2} \rvert \lt\varepsilon$
$\frac{x+2}{x^2+1} -\frac{3}{2}= \frac{-3x^2+2x+1}{2(x^2+1)}$
I have managed to express both the numerator and the denominator as such
$-3x^2+2x+1 = -(x-1)(3x+1)\\
2(x^2+1) = 2x^2+2= 2(x-1)^2 + 4(x-1) +4$
Returning back to the fraction $\frac{-(x-1)(3x+1)}{2(x-1)^2 + 4(x-1) +4}$

I am unsure on how to continue and would really appreciate some guidance.
 A: So you simplified the expression 
\begin{align}
\left | \frac{x+2}{x^2+1} -\frac{3}{2} \right | = \left | \frac{(x-1)(3x+1)}{2x^2+2} \right |.
\end{align}
The good news is we have an $|x-1|$ appearing, but there is also a $|3x+1|$ and $|2x^2+2|,$ which are expressions we don't necessarily want. The thing to do here is to bound these unwanted expressions by numbers, which I demonstrate in this problem.
Suppose $\delta < 1.$ What happens if $|x-1|< \delta$  ?
First, we see $0 < x < 2.$ Multiplying all sides of the inequality by $3$ and then adding $1$ to all sides, we see $1<3x+1<7,$ so $|3x+1|<7.$ Second, we see upon squaring all sides of $0<x<2$ and then adding $2$, $2<2x^2+2<10,$ so $\frac{1}{|2x^2+2|} < \frac{1}{2}.$
Now, for any $\epsilon>0,$ let us set $\delta= \min \lbrace 1, \frac{2 \epsilon}{7} \rbrace.$ Why does this work ?
If $\delta=1,$ then we see $|x-1| < \delta$ implies  $|x-1|<\frac{2 \epsilon}{7}$ and also
\begin{align}
\left|f(x)-\frac{3}{2} \right| = \left | \frac{x+2}{x^2+1} -\frac{3}{2} \right | = \left | \frac{(x-1)(3x+1)}{2x^2+2} \right |< \frac{7|x-1|}{2}< \frac{7}{2} \frac{2 \epsilon}{7}= \epsilon
\end{align} 
If $\delta=\frac{2 \epsilon}{7}$ then already we have $|x-1|< \delta$ implies
\begin{align}
\left|f(x)-\frac{3}{2} \right|=\left | \frac{x+2}{x^2+1} -\frac{3}{2} \right | = \left | \frac{(x-1)(3x+1)}{2x^2+2} \right |< \frac{7|x-1|}{2}< \frac{7}{2} \frac{2 \epsilon}{7}= \epsilon.
\end{align} 
A: Your fraction, in absolute value, equals
$$\left|\frac{(x-1)(3x+1)}{2(x^2+1)} \right|<\left|\frac{(x-1)(3x+1)}{x^2} \right|.$$
Now, if $|x-1|<\delta$, then $|x|<\delta+1$. So if $\delta\leq1$, then $|x|<2$ and $|x^2|<4$. So if we choose $\delta=\min\{1,\frac{7\epsilon}{4}\}$, then the above inequality is less than $\epsilon$, which is what we wanted to show.
A: Your inequality can be written
$$
\left|\frac{(x-1)(3x+1)}{x^2+1}\right|<2\varepsilon\tag{*}
$$
If $x>0$, then $x^2+1>1$, so $\frac{1}{x^2+1}<1$. The condition that $x>0$ is satisfied as soon as $\delta<1$, when you take $0<|x-1|<\delta$.
In this case, we have
$$
\left|\frac{(x-1)(3x+1)}{x^2+1}\right|<
\lvert(x-1)(3x+1)\rvert
$$
Since also $3x+1<7$, because $x<2$, we have
$$
\lvert(x-1)(3x+1)\rvert<7\lvert x-1\rvert<7\delta
$$
If we take $\delta=\min\{2\varepsilon/7,1\}$, we achieve what we want: for $0<|x-1|<\delta$, we have
$$
\left|\frac{(x-1)(3x+1)}{x^2+1}\right|<
\lvert(x-1)(3x+1)\rvert
<7\lvert x-1\rvert<7\delta\le7\frac{2\varepsilon}{7}=2\varepsilon
$$
The solution set of (*) may well be larger than the set described by
$$
0<\lvert x-1\rvert<\min\!\left\{\frac{2\varepsilon}{7},1\right\}
$$
but this is unimportant: what's needed for checking the correctness of a limit is that to every $\varepsilon>0$ there corresponds a $\delta>0$ et cetera: once you find a value for $\delta$ all numbers $\delta'$ with $0<\delta'<\delta$ would be good as well.
In some applications related to approximations, getting “the largest $\delta$” can be important, but not for the proof of correctness.
A: In this answer we continue where the OP left off, making one last 'normalization'  change:
$\tag 1 \frac{-(x-1)(3x+1)}{2(x-1)^2 + 4(x-1) +4} = \frac{-(x-1)(3(x-1)+4)}{2(x-1)^2 + 4(x-1) +4}$

Denominator:
If $|x - 1| \lt 1/2$ then
$\quad |2(x-1)^2 + 4(x-1) +4| \gt  |4(x-1) +4| \gt 2 $
Note that $4(x-1) \gt -2$.

$\text{ }$

Numerator:
If $|x - 1| \lt 1/2$ then  
$\quad \;  | -(x-1)\;(3\,(x-1)+4) | = |x-1| \, |3(x-1)+4| $
$ \qquad \quad \le | x-1|\;( 3 \,|x-1| + 4) \lt 5.5 \, |x-1| $
Use the triangle inequality.

Now the absolute value of the ratio expression (1) gets bigger as the numerator gets larger and the denominator gets smaller. So if $|x - 1| \lt 1/2$ and since $\frac{5.5}{2} = \frac{11}{4}$,
$\tag 2 \text{Absolute value of (1) } \lt \frac{11}{4} |x -1|$
We are now ready! Let the $\varepsilon \gt 0$ challenge be given and set
$\qquad \delta = min(\frac{1}{2},\frac{4 \varepsilon}{11})$
